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xxTIMURxx [149]
3 years ago
11

MoonieStardust Here is the rest of the picture from my last question you answered.

Computers and Technology
2 answers:
Oliga [24]3 years ago
6 0

Answer:

the two on the bottom and the two on the top

Explanation:

move the two on the bottom apart and the two on the top inward

Tasya [4]3 years ago
5 0
Sorry if I’m wrong but do the same thing as the last person I tried it and it’s right
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⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val
timurjin [86]

Answer:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

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